Answer:
The amount invested at 5% was $39,000 and the amount invested at 7% was $11,000
Step-by-step explanation:
we know that
The simple interest formula is equal to
where
I is the Final Interest Value
P is the Principal amount of money to be invested
r is the rate of interest
t is Number of Time Periods
Let
x-----> the amount invested at 5%
50,000-x -----> the amount invested at 7%
so
substitute in the formula above
solve for x
therefore
The amount invested at 5% was $39,000 and the amount invested at 7% was $11,000
7/10 of 80 is 56
4/5 of 80 is 64
3/4 of 80 is 60
5/8 of 80 is 50
1/2 of 80 is 40
1/4 of 80 is 20
1/10 of 80 is 8
1/8 of 80 is 10
to work these out divide the number by the denominator (bottom number) and multiply by the numerator (top number)
The picture in the attached figure
we know that
the area of the shaded region is equal to
(2/3)*[area of the circle - the area of the triangle]
step 1Find the area of a circle Ac
Ac = π r²
Ac = π (6)²
Ac = 113.10 units²
step 2find the area of the triangle At
The triangle is an equilateral triangle with angles on each corner equal to 60 degrees. Meanwhile,
the 3 angles at the center is 120 degrees each since a circle is 360 degree.
We know that the radius (line from centerpoint to corner) is equivalent to 6.
Using the cosine law,
we can calculate for the length of one side.
s² = 6^ + 6² – 2 (6) (6) cos 120
s² = 108
s = 10.4 units
Since this is an equilateral triangle, therefore, all sides are equal.
The area for this is:
At = (sqrt3 / 4) * s²
At = 46.77 units²
step 3the area of the shaded region=(2/3)*[area of the circle - the area of the triangle]
the area of the shaded region=(2/3)*[113.10-46.77]------> 44.22 units²
therefore
the answer isthe area of the shaded region is 44.22 units²
Answer:
5/9
Step-by-step explanation:
it's literally telling you the answer and asking you
If they want the odds in favor then it's the odds of the event happening
Answer:
x + 5
Step-by-step explanation:
Every time it goes up by a factor of 5
Tell me I am wrong.
Can I get brainliest
